Recent content by BruceG

Yes - start with Einstein, since QFT is just the quantization of special relativity. So with c=1, start with E^2 - p^2 = m^2 In terms of 4-momentum this is p^2 = m^2 In quantum theory with h= 1, p -> id/dx where x = (x,t) p = (p,E) So we get the wave equation...

Thanks for that. Now I have to work out if this argument is in someway equivalent to Baez' argument or provides an independent restriction. The trouble with a set like {1,2,4,8} is that pure coincidences can occur.

For some time I've been trying to get a geometric appreciation of why normed division algebras only exist in dimensions 1,2,4,8 (namely R,C,H,O). As always Baez provides the most elegant answer: http://math.ucr.edu/home/baez/octonions/node6.html" Allow me to descibe the key point of the...

I guess it is safe to say all that. Yes: the basic conflict of how to treat time, is that relativity demands we treat position and time on an equal footing as 4-vectors, however the canonical formulation of quantum mechanics requires we treat time separately. That's roughly it.

Yes it is a decent question. I've read through the suggested previous threads on the same topic and there are lots of good answers, but unfortunately no one knock-down answer. So we must end up staying a bit confused. The reason is that there are several levels of quantum formalism and we...

Here is a nice paper on this http://www.jyi.org/volumes/volume2/issue1/articles/barth.html" Basically the determinant is the general tool for calculation of a k-volume in k-space, but this paper explains doing k-volume in n-space when k<n.

Another way to say the same thing: imaginary numbers enter into quantum mechanics though the uncertainty principle: [X,P] = ih. In the wider context of particle physics we can say something like this: The geometrical settings for relativistic particle physics is 4 dimensional spacetime with...

In the many applications of inner product spaces e.g. functional analysis, quantum mechanics the underlying field just act as the scalars of the theory. Complex numbers are used as the default scalars because they are the unique complete algebraically-closed field so using C guarantees...

As a concrete example, it is interesting to see how the inner product is generalised to work over quaternions. First define conjugation by a+bi+cj+dk -> a-bi-cj-dk then the definition of inner product works much the same though you have to take a bit of care to deal with non-commutativity.

OK, ta, I got it. So with R and C the set of traceless matrices are closed (if a,b are traceless then so is [a,b]). So once you've counted the traceless matrices you've got the whole algebra. H and O are not commutative, so if you start with the set of traceless matrices, then to close off...

dimension of sl(2,R) = 1*(2*2-1) = 3, is isomorphic to so(2,1) : 2+1 = 3 dimension of sl(2,C) = 2*(2*2-1) = 6, is isomorphic to so(3,1) : 3+2+1 = 6 dimension of sl(2,H) = 15, is isomorphic to so(5,1) : 5+4+3+2+1 = 15 dimension of sl(2,O) = 45, is isomorphic to so(9,1) : 9+8+7+6+5+4+3+2+1 = 45...

This may be not the answer you are looking for, but you can do a proper generalisation to Rn by moving over to the wedge-product of 1-forms. Then the proof follows immediately for the assocoativity and antisymmetry of the wedge product: a^b^c = -a^c^b = +c^a^b Notes: 1. The special...

The analytic approach is: 1. Define exp(z) = 1+z+z^2/2!+... on C 2. Define ln(z) as the inverse exp(z) 3. Define a^z = exp(z.ln(a)) Then it follows that a^0 = exp(0) = 1.

As far as I'm aware the only solution is: n = 24 p = 2 C = 70

Also you're only going to get a local decent rate - decent rate will change as you move away from p.